
Originally Posted by
Jo37
Hello !
I'm a bit confused about the notion of Ker() and the solution I was given for an exercise.
f is a linear application defined for u = (x,y,z) by :
f(u) = (x + y + z, x + 2y + 3z, x + 3y + 5z)
Written as a system of equations :
$\displaystyle \left\{\begin{matrix} x + y + z = 0\\ x + 2y + 3z = 0\\ x + 3y + 5z = 0\end{matrix}\right.$
I was given as a solution Ker(f) = Vect{v1} = 1 with v1 =(-2,-3,1)
However, I have no idea how to get to this vector.
I solved the system by putting it in a rref matrix and got as solution x = z, y = -2z.
So this would give me v1 = (1,-2,1) if z equals one ? So I get Ker(f) = 1 also but I don't get why my vector is different and not even collinear to the one from the solution.
I also have trouble understanding how to interpret what Ker() or Im() represent (geometrically or just understand the concept).